The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant $K$ is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor. Select the correct statement(s)

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The force of attraction between the plates will become $K^{2}$ times the initial force

The electric field inside the capacitor will decrease $K$ times the initial field

The energy stored in the capacitor will become $K$ times the initial energy

The charge on the capacitor will become $K$ times the initial charge

answer is A, C, D.

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Detailed Solution

$(1) U = \frac{1}{2} C V^{2}$

As source between the plates is connected so potential difference remains constant. But capacitance $C$ becomes $K C$ hence energy stored is increased by factor $K$

$(2)$ electric field $\frac{V}{d}$ is not changed

$(3)$ Force of attraction F $\propto Q^{2} .$ Charge on each plate is increased by factor $K$ hence force between them increases by a factor of $K^{2}$

$(4)$ Since $Q = C V$ . Hence charge become $K Q$ as $C$ becomes $K C$ and $V$ remains unchanged